Linear Algebra and its Applications 306 (2000) 189–202
www.elsevier.com/locate/laa
Growth in Gaussian elimination for weighing
matrices, W (n, n − 1)
C. Koukouvinos a,∗ , M. Mitrouli b , Jennifer Seberry c
a Department of Mathematics, National Technical University of Athens, Zografou 15773, Athens, Greece
b Department of Mathematics, University of Athens, Panepistemiopolis 15784, Athens, Greece
c School of Information Technology and Computer Science, University of Wollongong, Wollongong, NSW
2522, Australia
Received 12 March 1999; accepted 8 November 1999
Submitted by R.A. Brualdi
Abstract
We consider the values for large minors of a skew-Hadamard matrix or conference matrix
W of order n and find that maximum n × n minor equals to (n − 1)n/2 , maximum
(n − 1) × (n − 1) minor equals to (n − 1)(n/2)−1 , maximum (n − 2) × (n − 2) minor equals
to 2(n − 1)(n/2)−2 , and maximum (n − 3) × (n − 3) minor equals to 4(n − 1)(n/2)−3 . This
leads us to conjecture that the growth factor for Gaussian elimination (GE) of completely
pivoted (CP) skew-Hadamard or conference matrices and indeed any CP weighing matrix of
order n and weight n − 1 is n − 1 and that the first and last few pivots are (1, 2, 2, 3 or 4, . . . ,
n − 1 or (n − 1)/2, (n − 1)/2, n − 1) for n > 14. We show the unique W (6, 5) has a single
pivot pattern and the unique W (8, 7) has at least two pivot structures. We give two pivot
patterns for the unique W (10, 9). © 2000 Elsevier Science Inc. All rights reserved.
AMS classification: 65F05; 65G05; 20B20
Keywords: Gaussian elimination; Growth; Complete pivoting; Weighing matrices
1. Introduction
Let A = [aij ] ∈ Rn×n . We reduce A to upper triangular form by using Gaussian
elimination (GE) operations. Let A(k) = [aij(k)] denote the matrix obtained after the
∗ Corresponding author. Fax: +30-1-772-1775.
E-mail address: [email protected] (C. Koukouvinos).
0024-3795/00/$ - see front matter 2000 Elsevier Science Inc. All rights reserved.
PII: S 0 0 2 4 - 3 7 9 5 ( 9 9 ) 0 0 2 5 4 - 2
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C. Koukouvinos et al. / Linear Algebra and its Applications 306 (2000) 189–202
first k pivoting operations, so A(n−1) is the final upper triangular matrix. A diagonal
entry of that final matrix will be called a pivot. Matrices with the property that no exchanges are actually needed during GE with complete pivoting are called completely
(k)
(0)
pivoted (CP) or feasible. Let g(n, A) = maxi,j,k |aij |/|a11 | denote the growth associated with GE on a CP A and g(n) = sup{ g(n, A)|A ∈ Rn×n }. The problem of
determining g(n) for various values of n is called the growth problem.
The determination of g(n) remains a mystery. Wilkinson [7] proved that
g(n) 6 [n 2 31/2 · · · n1/n−1 ]1/2 = f (n).
In Table 1, there are values of f (n) for representative values of n.
The above bound is certainly not sharp and the true upper bound is much smaller.
Wilkinson [8,9] noted that there were no known examples of matrices for which
g(n) > n. Cryer [2] conjectured that “g(n, A) 6 n, with equality iff A is a Hadamard
matrix”. This was proved to be untrue in [6].
A Hadamard matrix H of order n × n is an orthogonal matrix with elements ±1
and H H T = nI .
The problem is quite different if partial pivoting is allowed and Wilkinson [9, p.
212] gives an example of a matrix of order n and elements 0, ± 1 and growth factor
2n−1 .
It is easy to see that g(1) = 1 and g(2) = 2 for all n > 2. By using algebraic
methods, it was proved [1,2] that g(3) = 2.25, g(4) = 4 and g(5) 6 4 17
18 .
One of the curious frustrations of the growth problem is that it is quite difficult to
construct any examples of n × n matrices, A, other than Hadamard for which g(n, A)
is even close to n. Wilkinson [9] has remarked that in real-world problems, g(n, A)
has never been observed to be very large. Cryer [2] did numerical experiments in
which he computed g(n, A), doing complete pivoting on n × n matrices, A, with
entries chosen randomly from the interval [−1, 1] and for sizes up to n = 8. He had
to generate over 50,000 3 × 3 examples before finding one with g(3, A) > 2. Also
the largest g(n, A) he obtained by testing 10,000 random matrices for sizes up to
n = 8 was 2.8348.
Thus, in order to obtain matrices with large growth, sophisticated numerical optimization techniques must be applied [6]. By using such methods, matrices with
growth larger than n = 13, 14, 15, 16, 18, 20, 25 were specified, and thus the conjecture that g(n, A) 6 n is false. Table 2 summarizes the growth size attained for
various values of n [2,4].
The matrices that give rise to the growth factors of Table 2 are often extremely
sensitive to small perturbations in their entries in that tiny perturbations to a complete
Table 1
n
10
20
50
100
200
1000
f (n)
19
67
530
3300
26,000
7,900,000
C. Koukouvinos et al. / Linear Algebra and its Applications 306 (2000) 189–202
191
Table 2
n
13
14
15
16
18
20
25
Growth size
13.0205
14.5949
16.1078
18.0596
20.45
24.25
32.99
elimination matrix rarely results in another such matrix. This makes it rather difficult
to specify matrices which give rise to large growth.
If a Hadamard matrix, H, of order n can be written as H = I + S where S T = −S
then H is called skew-Hadamard.
A (0, 1, −1) matrix W = W (n, k) of order n satisfying W W T = kIn is called a
weighing matrix of order n and weight k or simply a weighing matrix. A W (n, n),
n ≡ 0 (mod 4), is a Hadamard matrix of order n. A W = W (n, k) for which W T =
−W is called a skew-weighing matrix. A W = W (n, n − 1) satisfying W T = W ,
n ≡ 2 (mod 4), is called a symmetric conference matrix. Conference matrices cannot exist unless n − 1 is the sum of two squares: thus they cannot exist for orders
22, 34, 58, 70, 78, 94. For more details and construction of weighing matrices the
reader can refer the book by Geramita and Seberry [5].
We have now studied, by computer, the pivots and growth factors for W (n, n −
1), n = 6, 10, 14, 18, 26, 30, 38, 42, 50, 54, 62, 74 constructed by two circulant
matrices and for n = 8, 12, 16, 20, 28, 36, 44, 52, 60, 68, 76, 84, 92, 100 constructed by four circulant matrices and obtained the results in Tables 3 and 4.
Wilkinson’s initial conjecture seems to be connected with Hadamard matrices.
Interesting results in the size of pivots appears when GE is applied to CP skewHadamard and weighing matrices of order n and weight n − 1. In these matrices, the
growth is also large, and experimentally, we have been led to believe it equals n − 1
and special structure appears for the first few and last few pivots. These results give
rise to new conjectures that can be posed for this category of matrices.
Conjecture (The growth conjecture for weighing matrices W (n, n − 1)). Let W =
W (n, n − 1) be a CP weighing matrix. Reduce W by GE. Then
(i) g(n, W ) = n − 1.
(ii) The three last pivots are equal to n − 1 or (n − 1)/2, (n − 1)/2, n − 1.
(iii) Every pivot before the last has magnitude at most n − 1.
(iv) The first four pivots are equal to 1, 2, 2, 3 or 4, for n > 14.
Notation. Write A for a matrix of order n whose initial pivots are derived from
matrices with CP structure. Write A(j ) for the absolute value of the determinant
of the j × j principal submatrix in the upper left-hand corner of the matrix A and
A[j ] for the absolute value of the determinant of the (n − j ) × (n − j ) principal
submatrix in the bottom right-hand corner of the matrix A. Throughout this paper
when we have used i pivots we then find all possible values of the A(n − i) minors.
Hence, if any minor is CP it must have one of these values. The magnitude of the
pivots appearing after the application of GE operations on a CP matrix W is given by
pj = W (j )/W (j − 1), j = 1, 2, . . . , n, W (0) = 1.
(1)
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We use W (j ), W [j ] similarly. We also use the following results
Lemma 1 [3]. Let A be an orthogonal matrix of order n satisfying AAT = kIn . Then
A(j ) = k j −(n/2) A[n − j ].
Corollary 1. If A is an n × n weighing matrix of weight k = n − 1, then the kth
pivot from the end is
pn+1−j =
kA[j − 1]
.
A[j ]
2. The first four pivots
Lemma 2. Let W be a CP weighing matrix, W (n, n − 1), of order n > 6 then if GE
is performed on W, the first three pivots are 1, 2, and 2.
Proof. We note that in the upper left-hand corner of a CP weighing matrix,
W (n, n − 1), of order n > 6 the following submatrices can always occur:
1 ,
1
1
1
,
−

1
1
1
1
−
1


1
1
1  or 1
1
−
1
−
1

1
0.
−
Thus, the first three pivots, using Eq. (1), are
p1 = 1,
p2 = 2
and p3 = 2.
Proposition 1. Let W be a CP weighing matrix, W (n, n − 1), of order n > 8. Then
if GE is performed on W, the first four pivots are 1, 2, 2, 3 or 4.
Proof. The first three pivots are given in Lemma 2. Now in the upper left-hand corner of a CP weighing matrix, W (n, n − 1), of order n > 8 the following submatrices
can always occur:




1 1 1 1
1 1 0 −
1 − 1 −
 1 − − −
1 − − 1  or 1 − 1 1  .
1 1 − −
1 1 − 1
The fourth pivots for n > 8, using Eq. (1), are
p4 = 4 or 3.
C. Koukouvinos et al. / Linear Algebra and its Applications 306 (2000) 189–202
193
3. Exact calculations
We assume that row and column permutations have been carried out. So we have
a CP skew-Hadamard or CP conference matrix A in the initial steps from which we
can calculate the maximum minors A(n), A(n − 1), A(n − 2) and A(n − 3). We
explore the use of a variation of a clever proof used by combinatorialists to find
the determinant of a matrix satisfying AAT = (k − λ)I + λJ , where I is the v × v
identity matrix, J the v × v matrix of ones and k, λ are integers to simplify our
proofs.
Proposition 2. Let A be a skew-Hadamard or conference matrix of order n. Then
the (n − 1) × (n − 1) minors are: A(n − 1) = 0, (n − 1)(n/2)−1.
Proof. Since AAT = (n − 1)I and det(A) = (n − 1)n/2 , the (n − 1) × (n − 1)
matrix B formed by deleting the first row and column of A satisfies det BB T =
(n − 1)n−2 or zero accordingly as the (1, 1) element of A is non-zero or zero. Hence
detB = (n − 1)(n/2)−1 or zero and we have the result. Proposition 3. Let A be a skew-Hadamard or conference matrix of order n. Then
the (n − 2) × (n − 2) minors are A(n − 2) = 0, 2(n − 1)(n/2)−2.
Proof. There are six cases: they have upper left-hand corner
1
1
1
1
1
0
1 1
,
,
or
.
1 −1
0 ±1
0 ±1
1 1
These have determinants 2, ± 1, ± 1, 0 respectively. We use the lower righthand principal minor, C, of order n − 2 to calculate CC T for each case. We find the
second case, where the determinant is (n − 1)(n/2)−2, is not CP as there
must be −2s after the first step of GE. Hence the maximum determinant of C is
2(n − 1)(n/2)−2. Lemma 3. The possible values for the determinants of 3 × 3 matrices with entries
0, ±1 where there is at most one zero in each row and column are 0, 1, ±2, ±3 and
±4.
Proof. For matrices of the required type, up to equivalence, we have these four cases

 
 
 

1
1
1
0
1
1
0
1
1
0
1
1
1 ±1 ±1 1 ±1 ±1 1
0 ±1 1
0 ±1 .
1 ±1 ±1
1 ±1 ±1
1 ±1 ±1
1 ±1
0
We used a computer to search all the possibilities and found that for no zeros the
determinant can be 0 or 4, for one zero the determinant can be 0, 2 or 4, for two
zeros the determinant can be 1 or 3, and for three zeros the determinant can be 0 or
2. 194
C. Koukouvinos et al. / Linear Algebra and its Applications 306 (2000) 189–202
We now proceed to study A(n − 3).
Proposition 4. Let A be a skew-Hadamard or conference matrix of order n. Then the
(n − 3) × (n − 3) minors are A(n − 3) = 0, 2(n − 1)(n/2)−3, or 4(n − 1)(n/2)−3
for n ≡ 0(mod 4) and 2(n − 1)(n/2)−3, or 4(n − 1)(n/2)−3 for n ≡ 2(mod 4).
Proof. We first note that the submatrices




1 1 1
1 1 1
1 − 1  and 1 − 0 
1 1 −
1 1 −
always occur in any skew-Hadamard or conference matrix of order > 6. We first
consider the upper left-hand corner


1 1 1
1 − 1  ,
1 1 −
which corresponds to a CP matrix with pivots 1, 2, 2.
We assume the CP matrix is in the form below where for ease of comprehension
we have written the elements a, b, c, d, p, q, s in the top 6 × 6 matrix although they
will not appear there in the CP matrix.

z
1 1 1 0 1 1 1

1 − 1 1 0 q 1

1 1 − p s 0 1

0 a c

1 0 b

1 d 0

1 1 1

 .. .. ..
. . .

1 1 1

1 1 −
C

 .. .. ..
. . .

1 1 −

1 − 1

. . .
 .. .. ..

1 − 1

1 − −

. . .
 .. .. ..
1 − −
u
}|
···
···
···
v
{ z }| {
1 1 ··· 1
1 1 ··· 1
1 − ··· −
w
z }| {
1 ··· 1
− ··· −
1 ··· 1

x
z }| {
1 · · · 1

− · · · −

− · · · −













.


















C. Koukouvinos et al. / Linear Algebra and its Applications 306 (2000) 189–202
195
We use the orthogonality of the matrix A, and the order n, to obtain constraints
for all the variables a, b, c, d, p, q, s, u, v, w, x in terms of each other and n, the
original order. We then calculate CC T and then these constraints are solved by either
Matlab or a simple, but tedious, calculation to obtain the values for the minors as 0
and 4(n − 1)(n/2)−3 for n ≡ 0(mod 4) and 4(n − 1)(n/2)−3 for n ≡ 2(mod 4).
We now consider the second case with upper left-hand corner,


1 1 1
1 − 0  ,
1 1 −
which also corresponds to a CP matrix with pivots 1, 2, 2.
We proceed, as before, to obtain the three values 0, 2(n − 1)(n/2)−3, 4(n −
(n/2)−3
for n ≡ 0(mod 4) and the two non-zero values 2(n − 1)(n/2)−3,
1)
(n/2)−3
as the only determinants for n ≡ 2(mod 4). 4(n − 1)
Theorem 1. When GE is applied on a CP skew-Hadamard or conference matrix
W of order n the last three pivots in backward order are n − 1, (n − 1)/2, and
(n − 1)/2 or n − 1.
Proof. The last three pivots are given by
W (n)
W (n − 1)
, pn−1 =
,
pn =
W (n − 1)
W (n − 2)
Since
pn−2 =
W (n − 2)
.
W (n − 3)
W (n) = (n − 1)n/2 ,
W (n − 1) = (n − 1)(n/2)−1 ,
W (n − 2) = 2(n − 1)(n/2)−2,
W (n − 3) = 2(n − 1)(n/2)−3
or 4(n − 1)(n/2)−3,
the values of the three last pivots are n − 1, (n − 1)/2, and (n − 1)/2 or n − 1,
respectively. 4. Numerical calculations
Lemma 4. The maximum determinant of all n × n matrices with elements ±1 or 0,
where there is at most one zero in each row and column is:
Order
Maximum
determinant
Possible determinantal values
2×2
3×3
4×4
5×5
2
4
16
48
0, 1, 2
0,1, 2, 3, 4
0,1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 16
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18,
19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 30, 32, 36, 40, 48
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C. Koukouvinos et al. / Linear Algebra and its Applications 306 (2000) 189–202
Remark. In fact, we found that considering all 5 × 5 matrices with elements ±1 and
no more than one zero per row and column, if the matrix had no zeros the determinant
could be 0, 16, 32 or 48; had exactly one zero the determinant could be 0, 8, 16, 24,
32 or 40; had exactly two zeros the determinant could be 0, 4, 8, 12, 16, 20, 24, 28,
32 or 36; had exactly three zeros the determinant could be 0, 2, 4, 6, 8, 10, 12, 14, 16,
18, 20, 22, 24, 26, 28, 30 or 36; had exactly four zeros the determinant could be 1,
3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25 or 27; had exactly five zeros the determinant
could be 0, 2, 4, 6, 8, 10, 12, 18, 20, 22.
Considering all 4 × 4 matrices with elements ±1 and no more than one zero per
row and column, if the matrix had no zeros the determinant could be 0, 8, 16; had
exactly one zero the determinant could be 0, 4, 8, 12; had exactly two zeros the
determinant could be 0, 2, 4, 6, 8, 10; had exactly three zeros the determinant could
be 1, 3, 5, 7, 9; had exactly four zeros the determinant could be 1, 3, 5, 9.
Considering all 3 × 3 matrices with elements ±1 and no more than one zero per
row and column, if the matrix had no zeros the determinant could be 0, 4; had exactly
one zero the determinant could be 0, 2, 4; had exactly two zeros the determinant
could be 1, 3; had exactly three zeros the determinant could be 0, 2.
Considering all 2 × 2 matrices with elements ±1 and no more than one zero per
row and column, if the matrix had no zeros the determinant could be 0, ±2; had
exactly one zero the determinant could be ±1; had exactly two zeros the determinant
could be ±1.
Lemma 5. W (4) = 10 for a W (6, 5).
Proof. Every 4 × 4 subdeterminant of W (6, 5) must contain two zeros. Hence, its
determinant can only be 0, 2, 4, 6, 8, or 10. We show that the first four non-zero
values are not possible in a W (6, 5).
Without any loss of generality we assume that the 4 × 4 subdeterminant has first
row and column comprising only +1s. Because we are dealing with a weighing
matrix the second row and column must contain two 1s and two −1s.
We denote the vectors (1, −, −), (1, −, 1) and (−, −, 1) as a1 , a2 and a3 , respectively. We denote the 2 × 2 submatrices
x 0
0 x
and
,
0 y
y 0
by b1 and b2 , respectively, where x and y are both 1 or −1.
Calculation shows that the 4 × 4 matrix with second row and column comprising
a1 and a1T can be completed by both b1 and b2 , but are equivalent under permutation
of rows and columns, to the matrix A1 below.
Furthermore calculations show that the 4 × 4 matrix with second row and column
ai and ajT , and completion matrix of shape bk , give only two inequivalent matrices,
A2 and A3 , under row and column permutations.
C. Koukouvinos et al. / Linear Algebra and its Applications 306 (2000) 189–202
(a2 , a2T , b1 ), (a3 , a3T , b1 )
and (a3 , a2T , b2 ) are equivalent to A1 .
(a2 , a2T , b2 ), (a3 , a3T , b2 )
and (a3 , a2T , b1 ) are equivalent to A2 .
Now, writing
we have

1
1

A1 = 
1
1
197
x for ±1, and using the orthogonality conditions for the W (6, 5),
1

1
1
1
−
− −

,
x 0
−
0
x̄
1


1
1

A2 = 
1
1
1
−
1
1
0
−

,
−
1
−
−
0
1
1
1
1


1
1

A3 = 
1
−
1
1 −

.
− 0
1
−
0
Now A1 has determinant 0. A2 and A3 have determinant 10. This gives the result.
Lemma 6. The unique pivots of the W (6, 5) are 1, 2, 2, 52 , 52 , 5 .
Proof. We use the determinants of W (1) = 1, W (2) = 2, W (3) = 4, W (4) = 10,
W [1] = 1, W [2] = 2.
Hence the pivot pattern is given by
W (2)
W (3)
= 2, p3 =
= 2,
p1 = 1, p2 =
W (1)
W (2)
5
W [1]
W [0]
= , p6 = 5
= 5.
W [2]
2
W [1]
o n
n
Lemma 7. The pivots of the W (8, 7) are 1, 2, 2, 4, 74 , 72 , 72 , 7 or 1, 2, 2, 3,
o
7 7 7
,
,
,
7
.
3 2 2
p4 =
5
W (4)
= ,
W (3)
2
p5 = 5
Proof. From Lemma 2 and Proposition 1, we have that
p1 = 1,
p2 = 2,
p3 = 2,
p4 = 4 or 3.
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C. Koukouvinos et al. / Linear Algebra and its Applications 306 (2000) 189–202
From Theorem 1, we also have that
7
7
and p6 = .
p8 = 7, p7 =
2
2
Since P8i=1 , pi = det W (8, 7) = 74 the only values that p5 can take are
7
4
or 73 .
Remark. The following matrices have pivot patterns 1, 2, 2, 4,
1, 2, 2, 3, 73 , 72 , 72 , 7, respectively.

1
1

1

1

1

1

0
1
1
−
−
1
−
1
1
0
1
−
1
−
0
−
1
1
1
1
−
−
1
0
1
−
1
1
−
0
−
−
−
1
1
−
0
1
1
−
−
−
1
0
1
−
−
1
−
−

0
−

−

−

1

1

−
1

1
1

1

1

0

1

1
1
1
−
−
1
1
1
−
0
0
−
1
−
1
1
1
−
−
−
1
1
1
−
0
1
−
1
0
−
1
1
−
1
1
1
1
0
1
−
−
−
1
0
−
−
1
−
1
1

−
1

−

1
.
1

0

1
−
7 7 7
4, 2, 2,
7 and
9 9 9
4, 2, 2,
o
9 or
and
n
Lemma 8. The pivots of the W (10, 9) can be 1, 2, 2, 3, 3, 4,
o
n
1, 2, 2, 4, 3, 3, 94 , 92 , 92 , 9 .
Proof. The W (10, 9) is unique up to permutation of rows and columns and multiplication of rows and columns by −1. We have found two CP W (10, 9) which have
difference pivot patterns showing the sensitivity of the pivots to permutations of rows
and columns.
o
n
The following matrices have pivot patterns 1, 2, 2, 3, 3, 4, 94 , 92 , 92 , 9 , and
o
n
1, 2, 2, 4, 3, 3, 94 , 92 , 92 , 9 , respectively.
C. Koukouvinos et al. / Linear Algebra and its Applications 306 (2000) 189–202

1
1

1

1

1


1

0

1

1
1
199

1
0

−

−

−


1

1

1

1
1
−
−
1
−
1
−
0
−
1
0
1
−
−
1
1
1
−
−
1
1
−
1
−
1
1
−
−
0
−
1
1
0
−
−
−
−
−
1
1
1
−
−
1
1
−
1
−
1
0
1
1
−
0
1
−
−
1
−
−
1
−
1
−
0
−
1
1
−
1
1
1
1
1
−
0
1
−
−
−
1
1
1
1
1
1
1
0
−
−
−
1
1
−
1
−
−
−
1
0
0
1
−
−
1
1
−
−
−
1
1
−
−
0
−
−
1
−
1
−
−
1
−
1
−
0
0
−
1
−
1
1
−
−
1
1
−
−
1
0
1
1
1
−
−
1
−
−
−
1
1
1
−


1

0


1
.
1


−

1


−
0
1
−
1
1
−
−
−
−
−
and

1
1


1

1


1

1


1

0


1
1
1

We calculated the values of all the large minors of the unique W (12, 11). These
are given in the next table. We also calculated all the minors for one of the W (20, 19)
and found exactly the same results as those in the table.
Minor
W (n − 1)
W (n − 2)
W (n − 3)
W (n − 4)
Minimum non-zero determinant
(n − 1)n/2−1
m=
m = (n − 1)n/2−2
m = (n − 1)n/2−3
m = (n − 1)n/2−4
All determinants
0, m
0, m, 2m
0, m, 2m, 3m, 4m
0, m, 2m, 3m, 4m, 6m, 8m,
9m, 10m, 12m, 16m
Tables 3 and 4 give us the pivot patterns calculated by a computer for the first
few W (n, n − 1) for both n ≡ 2(mod 4) and n ≡ 0(mod 4). Although our theory
predicts that the third last pivot could be n − 1 or (n − 1)/2, in both these tables
only the value (n − 1)/2 has been observed.
Growth
5
9
13
13
17
25
29
37
41
45
49
53
61
73
n
6
10
14
14
18
26
30
38
42
46
50
54
62
74
Table 3
Pivot pattern
1, 2, 2, 52 , 52 , 5
1, 2, 2, 3, 3, 4, 94 , 92 , 92 , 9 or 1, 2, 2, 4, 3, 3, 94 , 92 , 92 , 9
17
13 13 13 13
1, 2, 2, 3, 10
3 , 5 , 3.2941, 3.9464, 3.8235, 5/2 , 4 , 2 , 2 , 13
18
13 13 13 13 13 13
1, 2, 2, 3, 10
3 , 5 , 4, 4 , 3 , 3 , 4 , 2 , 2 , 13
18 11 148
17
17 17 17 17 17
1, 2, 2, 3, 10
3 , 5 , 3 , 33 , 3.7432, 4.5415, 3.9054, 17/5 , 10/3 , 3 , 4 , 2 , 2 , 17
18
25
13
1, 2, 2, 3, 10
3 , 5 , 4, 4, 4, 6 , 5, 3 , 6.7308, 5.2857, 4.3919, 4.9231, 5.9375,
25 , 25 , 25 , 25 , 25 , 25 , 25
6.0855, 5.5743, 11/3
3 3 4 2 2
18 34
29 29 29 29
1, 2, 2, 3, 10
3 , 5 , 9 , 3.9412, . . . , 3 , 4 , 2 , 2 , 29
18
13
37 37 37 37 37 37
1, 2, 2, 3, 10
3 , 5 , 4, 4, 3 , 4.4615, . . . , 4 , 8/3 , 3 , 4 , 2 , 2 , 37
18
37
41
41 41 41 41 41
1, 2, 2, 3, 10
3 , 5 , 4, 4, 9 , 4.4595, . . . , 8/3 , 10/3 , 3 , 4 , 2 , 2 , 41
18 34
45 45 45 45 45 45 45 45
1, 2, 2, 3, 10
3 , 5 , 9 , 4.2353, . . . , 4 , 6 , 4 , 4 , 2 , 4 , 2 , 2 , 45
18
40 9
49 49 49 49 49
1, 2, 2, 3, 10
3 , 5 , 4, 4, 9 , 2 , . . . , 8/3 , 3 , 4 , 2 , 2 , 49
18 34
53 53 53 53 53 53 53
1, 2, 2, 3, 10
3 , 5 , 9 , 4.4706, . . . , 6 , 4 , 4 , 2 , 4 , 2 , 2 , 53
18 34
37
61
61 61 61 61
1, 2, 2, 3, 10
3 , 5 , 9 , 4.2353, 9 , . . . , 16/5 , 5/2 , 4 , 2 , 2 , 61
18 , 4, 4, 37 , 4.9730, . . . , 73 , 73 , 73 , 73 , 73 , 73 , 73 , 73
,
1, 2, 2, 3, 10
3 5
9
6 4 4 2 4 2 2
200
C. Koukouvinos et al. / Linear Algebra and its Applications 306 (2000) 189–202
7
11
15
19
27
35
43
51
59
67
75
83
91
99
12
16
20
28
36
44
52
60
68
76
84
92
100
Growth
8
n
Table 4
19 , 19 , 19 , 19
4 2 2
11 , 11 , 11
2 2
27 , 27 , 27 , 27 , 27 , 27
5.5093, 6.0458, 5.4375, 6.5172, 6.6623, 6.0395, 7.9412, 10/3
3 4 2 2
18 34
35 35 35 35
1, 2, 2, 3, 10
3 , 5 , 9 , 3.8824, . . . , 3 , 4 , 2 , 2 , 35
18
37
43 43 43 43 43 43 43
1, 2, 2, 3, 10
3 , 5 , 4, 4, 9 , 5.0270, . . . , 6 , 4 , 4 , 2 , 4 , 2 , 2 , 43
18
37
51 51 51 51 51 51 51
1, 2, 2, 3, 10
3 , 5 , 4, 4, 9 , 4.7703, . . . , 25/4 , 4 , 4 , 2 , 4 , 2 , 2 , 51
18
38
59 59 59 59 59
1, 2, 2, 3, 10
3 , 5 , 4, 4, 9 , 5.1579, . . . , 10/3 , 3 , 4 , 2 , 2 , 59
18
38
67 67 67 67 67 67 67
1, 2, 2, 3, 10
3 , 5 , 4, 4, 9 , 5.1579, . . . , 6 , 4 , 4 , 2 , 4 , 2 , 2 , 67
18
40 26
75 75 75 75 75
1, 2, 2, 3, 10
3 , 5 , 4, 4, 9 , 5 , . . . , 10/3 , 3 , 4 , 2 , 2 , 75
18
37
83 83 83 83 83 83 83
1, 2, 2, 3, 10
3 , 5 , 4, 4, 9 , 5.1351, . . . , 6 , 4 , 4 , 2 , 4 , 2 , 2 , 83
18
40
41 91 91 91 91 91 91
1, 2, 2, 3, 10
3 , 5 , 4, 4, 9 , . . . , 6 , 4 , 4 , 2 , 4 , 2 , 2 , 91
18 , 4, 4, 38 , 5.4737, . . . , 99 , 99 , 99 , 99 , 99 , 99 , 99 , 99
,
1, 2, 2, 3, 10
3 5
9
6 4 4 2 4 2 2
Pivot pattern
1, 2, 2, 4, 74 , 72 , 72 , 7 or 1, 2, 2, 3, 73 , 72 , 72 , 7
17 11
11 11 11 11
10 11 18 11
1, 2, 2, 3, 10
3 , 5 , 17/5 , 5/2 , 4 , 2 , 2 , 11 or 1, 2, 2, 4, 3, 3 , 10/3 , 5 , 4 ,
18 31
15 15 15 15 15
1, 2, 2, 3, 10
3 , 5 , 9 , 4.0806, 3.913, 4.5455, 10/3 , 3 , 4 , 2 , 2 , 15
18 34
19 19
19 19
1, 2, 2, 3, 10
3 , 5 , 9 , 4.1176, 4.2857, 4.1, 4.6341, 25/6 , 4 , 5.2778, 10/3 , 3 ,
18 34
38
1, 2, 2, 3, 10
3 , 5 , 9 , 4.2353, 9 , 4, 4.9474, 4.6755, 4.9966, 4.8074, 5.5416,
C. Koukouvinos et al. / Linear Algebra and its Applications 306 (2000) 189–202
201
202
C. Koukouvinos et al. / Linear Algebra and its Applications 306 (2000) 189–202
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